Simplify the following expression and state the condition under which the simplification is valid. You can assume that $y \neq 0$. $t = \dfrac{6y}{5y^2 + y} \div \dfrac{-8}{4(5y + 1)} $
Answer: Dividing by an expression is the same as multiplying by its inverse. $t = \dfrac{6y}{5y^2 + y} \times \dfrac{4(5y + 1)}{-8} $ When multiplying fractions, we multiply the numerators and the denominators. $t = \dfrac{ 6y \times 4(5y + 1) } { (5y^2 + y) \times -8 } $ $ t = \dfrac {6y \times 4(5y + 1)} {-8 \times y(5y + 1)} $ $ t = \dfrac{24y(5y + 1)}{-8y(5y + 1)} $ We can cancel the $5y + 1$ so long as $5y + 1 \neq 0$ Therefore $y \neq -\dfrac{1}{5}$ $t = \dfrac{24y \cancel{(5y + 1})}{-8y \cancel{(5y + 1)}} = -\dfrac{24y}{8y} = -3 $